Question

I am pulling in weather data from a Yahoo weather RSS feed for both London and New York. I’m trying to reduce code duplication by reusing a PHP file which contains functions for pulling in the weather data.

Below is the function I am calling – get_current_weather_data(). This function goes off to various other functions such at the get_city() and get_temperature() functions.

<?php

function get_current_weather_data() {

// Get XML data from source

include_once 'index.php';  
$feed = file_get_contents(if (isset($sourceFeed)) { echo $sourceFeed; });

// Check to ensure the feed exists
if (!$feed) {
    die('Weather not found! Check feed URL');
}
$xml = new SimpleXmlElement($feed);

$weather = get_city($xml);
$weather = get_temperature($xml);
$weather = get_conditions($xml);
$weather = get_icon($xml);

return $weather;
}

In my index.php I set the URL for the RSS feed as the variable called $sourceFeed.

<?php

$tabTitle = ' | Home';
$pageIntroductionHeading = 'Welcome to our site';
$pageIntroductionContent = 'Twinz is a website which has been created to bring towns together! 
            Our goal is to bring communities around the world together, by providing 
            information about your home town and its twin town around the world. Our 
            site provides weather, news and background information for London and 
            one of its twin cities New York.'; 
$column1Heading = 'Current Weather for New York'; 
$column2Heading = 'Current Weather for London'; 
$column3Heading = 'Current Weather for Paris'; 

$sourceFeed = "http://weather.yahooapis.com/forecastrss?p=USNY0996&u=f"; 

include_once 'header.php';
include_once 'navigationMenu.php';
include_once 'threeColumnContainer.php';
include_once 'footer.php';

?>

I attempt to call the feed in my get_current_weather_data() function using:

(if (isset($sourceFeed)) { echo $sourceFeed; }).  

However, I receive the following error

"Warning: file_get_contents() [function.file-get-contents]: Filename cannot be empty in C:\xampp\htdocs\Twinz2\nyWeather.php on line 10
Weather not found! Check feed URL".

If I replace

(if (isset($sourceFeed)) { echo $sourceFeed; }) 

with the URL for the feed it works but this will stop me from reusing the code. Am I trying to do the impossible or is my syntax just incorrect?

This isset method works fine where used elsewhere like for example the $tabTitle and
$pageIntroductionHeading variables just need for the RSS feed.

Thanks in advance.

Answer 1

The problem is in the following line:

$feed = file_get_contents(if (isset($sourceFeed)) { echo $sourceFeed; });

it has to be:

if(isset($sourceFeed)){ $feed = file_get_contents($sourceFeed); }

And when you call the function, you also have to pass $sourceFeed as function parameter, like that:

get_current_weather_data($sourceFeed);